Functional equation with a parameter

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Functional equation with a parameter

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Source: Baltic Way 2024, Problem 1

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#1 h Nov 16, 2024, 4:56 PM • 1 Y

Y by EscoPablobar

Let $\alpha$ be a non-zero real number. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that
$xf(x+y)=(x+\alpha y)f(x)+xf(y)$ for all $x,y\in\mathbb{R}$ .

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pco

#2 h Nov 16, 2024, 5:22 PM

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Tintarn wrote:

Let $\alpha$ be a non-zero real number. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that
$xf(x+y)=(x+\alpha y)f(x)+xf(y)$ for all $x,y\in\mathbb{R}$ .

Let $P(x,y)$ be the assertion $xf(x+y)=(x+\alpha y)f(x)+xf(y)$

$P(1,0)$ $\implies$ $f(0)=0$

Let $x,y\ne 0$ :
$P(x,y)$ $\implies$ $f(x+y)=(1+\alpha \frac yx)f(x)+f(y)$
$P(y,x)$ $\implies$ $f(x+y)=(1+\alpha \frac xy)f(y)+f(x)$
Subtracting : $\frac yxf(x)=\frac xyf(y)$ , which is $\frac{f(x)}{x^2}=\frac{f(y)}{y^2}$ $\forall x,y\ne 0$
And so $f(x)=cx^2$ $\forall x\ne 0$ , still true when $x=0$

Plugging this back in original equation, we get any $c$ if $\alpha=2$ and $c=0$ if $\alpha\ne 2$

And so solutions :
$\forall \alpha$ : $\boxed{f(x)=0\quad\forall x}$ , which indeed fits
If $\alpha =2$ : $\boxed{f(x)=cx^2\quad\forall x}$ , which indeed fits, whatever is $c$

This post has been edited 1 time. Last edited by pco, Nov 16, 2024, 5:23 PM
Reason: Typos

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#3 h Nov 16, 2024, 9:18 PM • 1 Y

Y by NicoN9

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This post has been edited 1 time. Last edited by Flugan42, Nov 16, 2024, 9:19 PM
Reason: typo

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#4 h Nov 17, 2024, 11:29 AM

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Denote $P(x,y)$ the assertion of the given F.E.
$P(0,x)$ gives $\alpha xf(0)=0$ therefore $f(0)=0$ , now $P(x,-x)$ gives $xf(-x)=(\alpha-1)xf(x)$ so $f(-x)=(\alpha-1)f(x)$ for all $x \ne 0$ , now if $f(c)=0$ for $c \ne 0$ then $P(c,x)$ gives $f(c+x)=f(x)$ and then $P(x,c)$ gives $\alpha c f(x)=0$ which means $f(x)=0$ for all reals $d$ , so now if $f$ was injective at $0$ then swap $x$ with $-x$ to get $f(x)=(\alpha-1)f(-x)$ which gives $\alpha=2$ which means $f$ is even.
Now by $P(x,y)-P(x,-y)$ we get $x(f(x+y)-f(x-y))=4yf(x)$ so swapping $x,y$ gives that $\frac{4yf(x)}{x}=\frac{4xf(y)}{y}$ therefore $f(x)=c \cdot x^2$ for some constant $c$ and we can verify that any such $c$ work, thus we are done :cool:

:cool:

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#6 h Jan 7, 2025, 9:28 PM

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a short Sketch
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#7 h Jan 8, 2025, 7:06 PM

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Interesting!
We claim the only working function for $\alpha \ne 2$ is $f \equiv 0$ ; whereas for $\alpha = 2$ , the previously mentioned function works as does $f(x) \equiv cx^2$ for some $c \in \mathbb{R}$ . It can be seen that these solutions work, now we prove they're the only ones.

Let $P(x, y)$ denote the assertion. Since the only constant function is the zero function, assume $f$ is nonconstant.
$P(0, y) \implies f(0) = 0$ .
$P(x, -x)$ for $x \ne 0 \implies xf(-x) = (\alpha -1)xf(x) \implies f(-x) = (\alpha-1)f(x)$ $\implies f(x) = (\alpha-1)^2f(x) \implies \alpha = 2.$ (as $\alpha \ne 0$ )

Now we work on $x, y \ne 0$ .
Note that $(1+\frac{2x}{y})f(y)+f(x) = f(x+y) = (1+\frac{2y}{x})f(x) + f(y)$ $\implies \frac{f(x)}{x^2} = \frac{f(y)}{y^2},$ which gives the second curve of solutions described for $x \ne 0$ . But $f(0) = c*0^2$ , so we're done. :cool:

:cool:

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pie854

#8 h Feb 9, 2025, 4:59 PM

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Setting $y=0$ gives $f(0)=0$ . Note that, for $x\neq 0$ , $f(x+y)=f(x)+f(y)+\alpha \frac yx f(x).$ By symmetry, $\alpha \frac yx f(x)=\alpha \frac xy f(y)$ for all non-zero $x,y$ . Thus, $f(x)=x^2f(1)$ for all $x$ , which works.

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L13832

#9 h Feb 11, 2025, 4:15 AM • 1 Y

Y by alexanderhamilton124

Let $P(x,y)$ be the assertion $xf(x+y)=(x+\alpha y)f(x)+xf(y)$ .

$P(1,0)\implies f(1)=f(1)+f(0)\implies f(0)=0$ .

$P(x,y)\implies \left(1+\alpha\frac{y}{x}\right)f(x)+f(y)=f(x+y)=P(y,x)=\left(1+\alpha\frac{x}{y}\right)f(y)+f(x)$

$P(x,y)=P(y,x)\implies f(x)\frac{y}{x}=f(y)\frac{x}{y}$

If $y=1$ we get $f(x)=f(1)x^2=cx^2$ . Plugging this into the equation we see that $\boxed{f\equiv 0}$ is also a solution and $\boxed{f(x)=cx^2}$ is a solution for $\alpha = 2$ .

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NicoN9

#10 h Yesterday at 11:58 AM

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We claim that the solution is $f(x)\equiv 0$ , and $f(x)\equiv cx^2$ when $\alpha =2$ . we easily see that these satisfys the condition. Let $P(x, y)$ denotes the assertion.

First, by $P(x, 0)$ we easily see that $f(0)=0$ . We casework on $\alpha$ .

case 1: $\alpha \neq 1, 2$ .
From $P(x, -x)$ , and $P(-x, x)$ , we confirm that $\begin{align*} x(\alpha -1)f(x) &= xf(-x)\\ &= x\frac{xf(x)}{x(\alpha -1)}\\ &= \frac{xf(x)}{\alpha -1}. \end{align*}$ Suppose $f(x)\neq 0$ and $x\neq 0$ . Now from above, $\alpha -1 =\frac{1}{\alpha - 1}$ which is impossible.

case 2: $\alpha =1$ .
In this case, just check $P(x, -x)$ to get $f(x)=0$ .

case 3: $\alpha =2$ .
Compare $P(1, x)$ and $P(x, 1)$ , we have $(x+2)f(x)+xf(1)=x(2x+1)f(1)+xf(x)$ and this is equivalent to $f(x)=f(1)x^2$ .

And as these cover all the cases, we are done.

This post has been edited 5 times. Last edited by NicoN9, Yesterday at 10:15 PM
Reason: Too much error + English.

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